CHEMISTRY · LESSON 02 OF 12
Quantum theory
Light comes in photons with energy E = hν. Electrons in atoms can only have certain energies, so atoms absorb and emit light only at particular wavelengths. Electrons are described by orbitals labelled by quantum numbers.
What this lesson explains
Quantum theory explains why elements have their chemical properties, why they give characteristic colours in flames and spectra, and how spectroscopy — the main analytical tool of chemistry — identifies substances and measures concentrations. It also underlies lasers, LEDs, solar cells and the periodic table itself.
Before you begin
Scientific notation
Quantum quantities are tiny: h = 6.626 × 10−34 J·s. Multiply coefficients and add exponents: (6.626 × 10−34)(2.998 × 108) = 1.986 × 10−25.
Unit conversions
1 nm = 10−9 m. Energy per mole = energy per photon × NA (6.022 × 1023 mol−1).
The idea, made visible
Light is an electromagnetic wave with wavelength λ and frequency ν related by c = λν. But in 1900–1905 Planck and Einstein showed that light energy comes in packets called photons, each with energy E = hν = hc / λ. Shorter wavelength means higher energy: ultraviolet photons carry more energy than red ones.
The photoelectric effect proved this. Light ejects electrons from a metal only if each photon has enough energy (frequency above a threshold). Brighter light of too low a frequency ejects none, because brightness means more photons, not more energetic ones.
Heated hydrogen gas emits light only at certain wavelengths — a line spectrum, not a continuous rainbow. Bohr explained this: the electron can only occupy levels with energies En = −2.18 × 10−18 J / n2, n = 1, 2, 3, … When it drops from a higher level to a lower one, it emits a photon whose energy equals the difference, ΔE = hν. The energies are negative because the electron is bound; zero means the electron has been removed (ionised).
Electrons also behave as waves (de Broglie, λ = h/mv), and the full quantum-mechanical model replaces Bohr’s orbits by orbitals: regions where the electron is likely to be found. Each orbital is labelled by three quantum numbers, and each electron by a fourth: n (shell, size and energy), l (subshell shape: s, p, d, f), ml (orientation) and ms (spin, +½ or −½).
The Heisenberg uncertainty principle says an electron’s position and momentum cannot both be known exactly, which is why we speak of probability, not paths.
| n | l (subshell) | ml values | orbitals | max electrons |
|---|---|---|---|---|
| 1 | 0 (1s) | 0 | 1 | 2 |
| 2 | 0 (2s), 1 (2p) | 0; −1, 0, +1 | 1 + 3 = 4 | 8 |
| 3 | 0 (3s), 1 (3p), 2 (3d) | 0; −1…+1; −2…+2 | 1 + 3 + 5 = 9 | 18 |
There is no 2d or 1p subshell, because l must be less than n.
Key terms
- Wavelength λ, frequency ν
- λ: distance between wave crests (m); ν: number of waves per second (Hz = s−1).
- Photon
- A quantum (packet) of light energy, E = hν.
- Ground and excited states
- The lowest-energy arrangement (n = 1 for hydrogen) and any higher-energy arrangement.
- Emission / absorption
- An electron dropping to a lower level emits a photon; a photon of exactly the right energy can raise an electron to a higher level.
- Orbital
- A region of space described by a wavefunction where an electron of given energy is likely to be found; it holds at most two electrons of opposite spin.
- Quantum numbers
- n = 1, 2, 3, …; l = 0 to n − 1 (s, p, d, f for l = 0, 1, 2, 3); ml = −l to +l; ms = ±½.
The formulas and what they mean
| Symbol | Meaning | Unit |
|---|---|---|
| h | Planck’s constant, 6.626 × 10⁻³⁴ | J·s |
| c | speed of light, 2.998 × 10⁸ | m/s |
| λ | wavelength | m (often nm) |
| ν | frequency | Hz = s⁻¹ |
| En | energy of level n (hydrogen) | J |
| n, l, ml, ms | quantum numbers | — |
Wave relation
c = λν
Conditions and limits: All electromagnetic radiation in vacuum.
Photon energy
E = hν = hc / λ
Conditions and limits: Energy of one photon. Multiply by NA for a mole of photons.
Hydrogen energy levels (Bohr)
En = −2.18 × 10−18 J / n2
Conditions and limits: Hydrogen atom (one electron) only. Multi-electron atoms need the full quantum model.
Transition energy
ΔE = 2.18 × 10−18 J (1 / nf2 − 1 / ni2) ; |ΔE| = hc / λ
Conditions and limits: Hydrogen. ΔE < 0 for emission (n_i > n_f).
de Broglie wavelength
λ = h / mv
Conditions and limits: Any moving particle; significant only for very small masses such as electrons.
Orbital counts
n2 orbitals and 2n2 electrons per shell; 2l + 1 orbitals per subshell
Conditions and limits: s: 1 orbital, p: 3, d: 5, f: 7.
Why it works: The red line of hydrogen from the energy levels
Predict the wavelength emitted when an electron drops from n = 3 to n = 2.
ΔE = 2.18 × 10−18(1 / 4 − 1 / 9) = 2.18 × 10−18 × 0.1389 = 3.03 × 10−19 J.
Energy released (magnitude).
λ = hc / ΔE = (6.626 × 10−34)(2.998 × 108) / 3.03 × 10−19 = 6.56 × 10−7 m.
Photon energy relation.
= 656 nm, red light.
1 nm = 10⁻⁹ m.
This matches the measured red line of hydrogen (656.3 nm) — strong evidence that energies are quantised.
A first worked example
Photon and spectrum calculations
- Convert wavelengths to metres (nm × 10⁻⁹).
- Use c = λν to switch between wavelength and frequency.
- Use E = hν = hc/λ for one photon; multiply by NA and divide by 1000 for kJ/mol.
- For hydrogen transitions, compute ΔE from the level formula, then λ = hc/|ΔE|.
- Check the region: 400–700 nm is visible; shorter is UV, longer is IR.
Energy of a photon
Problem. Find the energy of one photon of green light (λ = 500 nm) and of one mole of such photons.
E = hc / λ = (6.626 × 10−34)(2.998 × 108) / 500 × 10−9 = 3.97 × 10−19 J.
Convert nm to m first.
Per mole: 3.97 × 10−19 × 6.022 × 1023 = 2.39 × 105 J/mol = 239 kJ/mol.
Multiply by Avogadro’s number.
Result: 3.97 × 10−19 J per photon; 239 kJ/mol.
What it means: This is comparable to chemical bond energies, which is why visible and UV light can drive chemical reactions (photosynthesis, fading of dyes).
A different case
Scaling with wavelength
Problem. A photon’s wavelength is halved. By what factor does its energy change? If its frequency is tripled instead?
E = hc/λ: halving λ doubles E.
Energy is inversely proportional to wavelength.
E = hν: tripling ν triples E.
Energy is proportional to frequency.
Result: ×2 and ×3.
What it means: Blue photons (≈ 450 nm) carry about 1.5 times the energy of red ones (≈ 680 nm).
More worked cases
Each case below uses a different skill. Every step and result is shown.
Frequency from wavelength
Problem. A radio station broadcasts at 100 MHz. Find the wavelength.
λ = c / ν = 2.998 × 108 / 1.00 × 108 = 3.00 m.
c = λν.
Result: λ ≈ 3.0 m.
What it means: Radio photons carry very little energy (6.6 × 10⁻²⁶ J): far too little to break bonds.
Ionisation energy of hydrogen
Problem. How much energy is needed to remove the electron from a ground-state hydrogen atom? Express it per mole.
From n = 1 to n = ∞: ΔE = 0 − (−2.18 × 10−18) = 2.18 × 10−18 J.
E_∞ = 0.
× 6.022 × 1023 = 1.31 × 106 J/mol = 1312 kJ/mol.
Per mole.
Result: 2.18 × 10−18 J per atom = 1312 kJ/mol.
What it means: This agrees with the measured first ionisation energy of hydrogen.
Allowed quantum numbers
Problem. Which set is not allowed: (a) n = 2, l = 1, ml = −1; (b) n = 3, l = 3, ml = 0; (c) n = 4, l = 2, ml = +2?
(a) l = 1 < 2 and |ml| ≤ 1: allowed (a 2p orbital).
Check l < n and −l ≤ m_l ≤ l.
(b) l = 3 is not less than n = 3: not allowed.
l can be at most n − 1 = 2.
(c) l = 2 < 4 and ml = +2 ≤ 2: allowed (a 4d orbital).
Both rules satisfied.
Result: Set (b) is not allowed.
What it means: This is why there are no “3f” orbitals.
Common misunderstandings
Misunderstanding: Forgetting to convert nm to m.
Correct idea: Multiply nanometres by 10⁻⁹ before using c or hc.
Misunderstanding: “Brighter light has more energetic photons.”
Correct idea: Brightness is the number of photons; each photon’s energy depends only on its frequency.
Misunderstanding: Applying E_n = −2.18 × 10⁻¹⁸/n² J to atoms other than hydrogen.
Correct idea: The Bohr formula works for one-electron species only.
Misunderstanding: Thinking of orbitals as fixed circular paths.
Correct idea: Orbitals are probability regions; the electron has no definite path.
Misunderstanding: Allowing l = n.
Correct idea: l ranges from 0 to n − 1.
Keep in mind
- c = λνWave relation
- E = hν = hc / λPhoton energy
- En = −2.18 × 10−18 J / n2Hydrogen energy levels (Bohr)
- ΔE = 2.18 × 10−18 J (1 / nf2 − 1 / ni2) ; |ΔE| = hc / λTransition energy
- λ = h / mvde Broglie wavelength
- n2 orbitals and 2n2 electrons per shell; 2l + 1 orbitals per subshellOrbital counts
Scope of this lesson
- Wavefunction mathematics (Schrödinger equation) is described qualitatively only.
- Electron configurations and the filling order are developed in the Periodic table topic.
Next: Periodic table. Electron configurations follow from filling orbitals in order of energy. The periodic table groups elements with similar outer electrons, and effective nuclear charge explains the trends in size, ionisation energy and electronegativity.
Original study text. Sources and credits.